Question

Let $A$ and $B$ (where $A>B$), be acute angles. If $\sin (A+B)=\frac{12}{13}$ and $\cos (A-B)=\frac{3}{5}$, then the value(s) of $\sin \left(2A\right)$ is/are
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• A. $\frac{56}{65}$
• B. $\frac{16}{65}$
• C. $-\frac{56}{65}$
• D. $-\frac{16}{65}$

Answer 1

Answer: (A), (B)

$\frac{16}{65}$

$\sin (A+B)=\frac{12}{13}$
So, $A+B$ lies in first or second quadrant,
$\Rightarrow \cos (A+B)=\pm \frac{5}{13}$
$\cos (A-B)=\frac{3}{5}$
$\because A,B$ are acute angles and $A>B$$\Rightarrow A-B$ lies in the first quadrant,
$\Rightarrow \sin (A-B)=\frac{4}{5}$
Now,
$\sin \left(2A\right)=\sin (A+B+A-B)=\sin (A+B)\cos (A-B)+\cos (A+B)\sin (A-B)=\frac{12}{13}\times \frac{3}{5}\pm \frac{5}{13}\times \frac{4}{5}=\frac{36}{65}+\frac{20}{65}\text{or}\frac{36}{65}-\frac{20}{65}=\frac{56}{65}\text{or}\frac{16}{65}$